For positive integers $m$ and $v$, let $$b_5(m,v) := \min \{ a_m : a \in \mathbb N^m, 1 \le a_1 < \cdots < a_m, \ v_5(a_1 \cdot\ldots\cdot a_m) \ge v \}.$$ The values of $b_5(m,v)$ that we will use in the following can easily be determined by computer.

Let $n \ge 2$ be an integer and $x \in \mathbb N^n$ a solution, i.e. $1 \le x_1 < \cdots < x_n$ with $$x_1 \cdot\ldots\cdot x_n = 100^{n-1} (x_1 + \cdots + x_n),$$ which implies $x_n \ge b_5(n, 2n-2))$.

Following Servaes' idea, we estimate $$\begin{eqnarray*} (n-1)! \cdot x_n &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1}(n-1)x_{n-1} + 100^{n-1}x_n \end{eqnarray*}$$ and get $$x_{n-1} \ge \alpha_n \, x_n \quad \mbox{with} \quad \alpha_n := \frac{(n-2)!}{100^{n-1}} - \frac1{n-1}.$$ In the following, let $n \ge 270$, which ensures $\alpha_n > 0$. Then we have: $$\begin{eqnarray*} (n-2)! \cdot x_{n-1} \cdot x_n &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} n x_n \end{eqnarray*}$$ $$(n-2)! \cdot \alpha_n \, x_n \le (n-2)! \cdot x_{n-1} \le 100^{n-1} n$$ $$x_n \le \frac{100^{n-1}n}{(n-2)! \cdot \alpha_n} =: \beta_n \tag{1}$$ Furthermore we have: $$\begin{eqnarray} (n-2)! \cdot x_{n-1} \cdot (x_{n-1} + 1) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1} (x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-1) x_{n-1} + 100^{n-1} \cdot \beta_n \end{eqnarray} \tag{2}$$ $$\begin{eqnarray} (n-3)! \cdot x_{n-2} (x_{n-2} + 1) (x_{n-2} + 2) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-2) x_{n-2} + 100^{n-1} \cdot 2 \beta_n \end{eqnarray} \tag{3}$$ $$\begin{eqnarray} (n-4)! \cdot x_{n-3} (x_{n-3} + 1) (x_{n-3} + 2) (x_{n-3} + 3) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-3) x_{n-3} + 100^{n-1} \cdot 3 \beta_n \end{eqnarray} \tag{4}$$ In the case of $n = 274$, $x_n \ge b_5(274, 546) = 5500$. But (1) gives $x_n \le 1144$, a contradiction.

In the case of $n = 273$, (1) reads $x_n \le 8549$, so $v_5(x_n) \le 5$ and $$544 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-1}) + 5,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-1}) \ge 539$, so $x_{n-1} \ge b_5(272,539) = 5375$. But (2) gives $x_{n-1} \le 1536$, a contradiction.

In the case of $n = 272$, (1) reads $x_n \le 64877$, so $v_5(x_n) \le 6$ and $$542 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-1}) + 6,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-1}) \ge 536$, so $x_{n-1} \ge b_5(271,536) = 5350$. But (2) gives $x_{n-1} \le 4293$, a contradiction.

In the case of $n = 271$, (1) reads $x_n \le 523728$, so $v_5(x_n) \le 8$, and (2) yields $x_{n-1} \le 12622$, so $v_5(x_{n-1}) \le 5$ and $$540 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-2}) + 13,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-2}) \ge 527$, so $x_{n-2} \ge b_5(269,527) = 5200$. But (3) gives $x_{n-2} \le 2680$, a contradiction.

In the case of $n = 270$, (1) reads $x_n \le 5395811$, so $v_5(x_n) \le 9$, (2) yields $x_{n-1} \le 43004$, so $v_5(x_{n-1}) \le 6$, and (3) yields $x_{n-2} \le 7188$, so $v_5(x_{n-2}) \le 5$ and $$538 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-3}) + 20,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-3}) \ge 518$, so $x_{n-3} \ge b_5(267,518) = 5050$. But (4) gives $x_{n-3} \le 3396$, a contradiction.

Consequently, $n \le 269$.

For positive integers $m$ and $v$, let $$b_5(m,v) := \min \{ a_m : a \in \mathbb N^m, 1 \le a_1 < \cdots < a_m, \ v_5(a_1 \cdot\ldots\cdot a_m) \ge v \}.$$ The values of $b_5(m,v)$ that we will use in the following can easily be determined by computer.

Let $n \ge 2$ be an integer and $x \in \mathbb N^n$ a solution, i.e. $1 \le x_1 < \cdots < x_n$ with $$x_1 \cdot\ldots\cdot x_n = 100^{n-1} (x_1 + \cdots + x_n),$$ which implies $x_n \ge b_5(n, 2n-2)$.

Following Servaes' idea, we estimate $$\begin{eqnarray*} (n-1)! \cdot x_n &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1}(n-1)x_{n-1} + 100^{n-1}x_n \end{eqnarray*}$$ and get $$x_{n-1} \ge \alpha_n \, x_n \quad \mbox{with} \quad \alpha_n := \frac{(n-2)!}{100^{n-1}} - \frac1{n-1}.$$ In the following, let $n \ge 270$, which ensures $\alpha_n > 0$. Then we have: $$\begin{eqnarray*} (n-2)! \cdot x_{n-1} \cdot x_n &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} n x_n \end{eqnarray*}$$ $$(n-2)! \cdot \alpha_n \, x_n \le (n-2)! \cdot x_{n-1} \le 100^{n-1} n$$ $$x_n \le \frac{100^{n-1}n}{(n-2)! \cdot \alpha_n} =: \beta_n \tag{1}$$ Furthermore we have: $$\begin{eqnarray} (n-2)! \cdot x_{n-1} \cdot (x_{n-1} + 1) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1} (x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-1) x_{n-1} + 100^{n-1} \cdot \beta_n \end{eqnarray} \tag{2}$$ $$\begin{eqnarray} (n-3)! \cdot x_{n-2} (x_{n-2} + 1) (x_{n-2} + 2) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-2) x_{n-2} + 100^{n-1} \cdot 2 \beta_n \end{eqnarray} \tag{3}$$ $$\begin{eqnarray} (n-4)! \cdot x_{n-3} (x_{n-3} + 1) (x_{n-3} + 2) (x_{n-3} + 3) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-3) x_{n-3} + 100^{n-1} \cdot 3 \beta_n \end{eqnarray} \tag{4}$$ In the case of $n = 274$, $x_n \ge b_5(274, 546) = 5500$. But (1) gives $x_n \le 1144$, a contradiction.

In the case of $n = 273$, (1) reads $x_n \le 8549$, so $v_5(x_n) \le 5$ and $$544 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-1}) + 5,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-1}) \ge 539$, so $x_{n-1} \ge b_5(272,539) = 5375$. But (2) gives $x_{n-1} \le 1536$, a contradiction.

In the case of $n = 272$, (1) reads $x_n \le 64877$, so $v_5(x_n) \le 6$ and $$542 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-1}) + 6,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-1}) \ge 536$, so $x_{n-1} \ge b_5(271,536) = 5350$. But (2) gives $x_{n-1} \le 4293$, a contradiction.

In the case of $n = 271$, (1) reads $x_n \le 523728$, so $v_5(x_n) \le 8$, and (2) yields $x_{n-1} \le 12622$, so $v_5(x_{n-1}) \le 5$ and $$540 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-2}) + 13,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-2}) \ge 527$, so $x_{n-2} \ge b_5(269,527) = 5200$. But (3) gives $x_{n-2} \le 2680$, a contradiction.

In the case of $n = 270$, (1) reads $x_n \le 5395811$, so $v_5(x_n) \le 9$, (2) yields $x_{n-1} \le 43004$, so $v_5(x_{n-1}) \le 6$, and (3) yields $x_{n-2} \le 7188$, so $v_5(x_{n-2}) \le 5$ and $$538 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-3}) + 20,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-3}) \ge 518$, so $x_{n-3} \ge b_5(267,518) = 5050$. But (4) gives $x_{n-3} \le 3396$, a contradiction.

Consequently, $n \le 269$.

For positive integers $k$ and $m$, let $b_5(k,m)$ be the smallest positive integer for which there exist positive integers $a_1 < \cdots < a_k \le b_5(k,m)$ with $v_5(a_1 \cdots a_k) \ge m$. The values $b_5(273,544) = 5475$, $b_5(272,542) = 5450$, $b_5(271,536) = 5350$ can easily be determined by computer.

Assuming $n = 273$ and a solution $x_1 < \cdots < x_n$, i.e. $x_1 \cdots x_n = 100^{n-1}(x_1 + \cdots + x_n)$, Servaes' calculation gives $x_n \le 8549$. Sadly, $v_5(8549!)$ is too large to use it in Servaes' way, but $x_n \ge b_5(273,544) = 5475$ gives

$$(n-2)! \cdot x_{n-1} \cdot 5475 \le x_1 \cdots x_n = 100^{n-1} (x_1 + \cdots + x_n) \le 100^{n-1}(n-1) x_{n-1} + 100^{n-1} \cdot 8549,$$

which requires $x_{n-1} \le 11$. This is not possible.

So we get $n \le 272$.

For the case of $n = 272$, the same arguments give $5450 \le x_n \le 64877$ and $x_{n-1} \le 704$. Because $v_5(x_n) \le 6$, we would need positive integers $x_1 < \dots < x_{n-1} \le 704$ with $v_5(x_1 \cdots x_{n-1}) \ge 542 - 6 = 536$. But $b_5(271,536) = 5350 > 704$, so again, this cannot be.

Consequently, $n \le 271$.

For positive integers $m$ and $v$, let $$b_5(m,v) := \min \{ a_m : a \in \mathbb N^m, 1 \le a_1 < \cdots < a_m, \ v_5(a_1 \cdot\ldots\cdot a_m) \ge v \}.$$ The values of $b_5(m,v)$ that we will use in the following can easily be determined by computer.

Let $n \ge 2$ be an integer and $x \in \mathbb N^n$ a solution, i.e. $1 \le x_1 < \cdots < x_n$ with $$x_1 \cdot\ldots\cdot x_n = 100^{n-1} (x_1 + \cdots + x_n),$$ which implies $x_n \ge b_5(n, 2n-2))$.

Following Servaes' idea, we estimate $$\begin{eqnarray*} (n-1)! \cdot x_n &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1}(n-1)x_{n-1} + 100^{n-1}x_n \end{eqnarray*}$$ and get $$x_{n-1} \ge \alpha_n \, x_n \quad \mbox{with} \quad \alpha_n := \frac{(n-2)!}{100^{n-1}} - \frac1{n-1}.$$ In the following, let $n \ge 270$, which ensures $\alpha_n > 0$. Then we have: $$\begin{eqnarray*} (n-2)! \cdot x_{n-1} \cdot x_n &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} n x_n \end{eqnarray*}$$ $$(n-2)! \cdot \alpha_n \, x_n \le (n-2)! \cdot x_{n-1} \le 100^{n-1} n$$ $$x_n \le \frac{100^{n-1}n}{(n-2)! \cdot \alpha_n} =: \beta_n \tag{1}$$ Furthermore we have: $$\begin{eqnarray} (n-2)! \cdot x_{n-1} \cdot (x_{n-1} + 1) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1} (x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-1) x_{n-1} + 100^{n-1} \cdot \beta_n \end{eqnarray} \tag{2}$$ $$\begin{eqnarray} (n-3)! \cdot x_{n-2} (x_{n-2} + 1) (x_{n-2} + 2) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-2) x_{n-2} + 100^{n-1} \cdot 2 \beta_n \end{eqnarray} \tag{3}$$ $$\begin{eqnarray} (n-4)! \cdot x_{n-3} (x_{n-3} + 1) (x_{n-3} + 2) (x_{n-3} + 3) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-3) x_{n-3} + 100^{n-1} \cdot 3 \beta_n \end{eqnarray} \tag{4}$$ In the case of $n = 274$, $x_n \ge b_5(274, 546) = 5500$. But (1) gives $x_n \le 1144$, a contradiction.

In the case of $n = 273$, (1) reads $x_n \le 8549$, so $v_5(x_n) \le 5$ and $$544 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-1}) + 5,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-1}) \ge 539$, so $x_{n-1} \ge b_5(272,539) = 5375$. But (2) gives $x_{n-1} \le 1536$, a contradiction.

In the case of $n = 272$, (1) reads $x_n \le 64877$, so $v_5(x_n) \le 6$ and $$542 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-1}) + 6,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-1}) \ge 536$, so $x_{n-1} \ge b_5(271,536) = 5350$. But (2) gives $x_{n-1} \le 4293$, a contradiction.

In the case of $n = 271$, (1) reads $x_n \le 523728$, so $v_5(x_n) \le 8$, and (2) yields $x_{n-1} \le 12622$, so $v_5(x_{n-1}) \le 5$ and $$540 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-2}) + 13,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-2}) \ge 527$, so $x_{n-2} \ge b_5(269,527) = 5200$. But (3) gives $x_{n-2} \le 2680$, a contradiction.

In the case of $n = 270$, (1) reads $x_n \le 5395811$, so $v_5(x_n) \le 9$, (2) yields $x_{n-1} \le 43004$, so $v_5(x_{n-1}) \le 6$, and (3) yields $x_{n-2} \le 7188$, so $v_5(x_{n-2}) \le 5$ and $$538 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-3}) + 20,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-3}) \ge 518$, so $x_{n-3} \ge b_5(267,518) = 5050$. But (4) gives $x_{n-3} \le 3396$, a contradiction.

Consequently, $n \le 269$.